For real spaces this is clear because identity is the only nonzero field homomorphism of \(\mathbb {R}\) . In the sequel we assume \(X,Y\)  are complex normed spaces, so \(\mathbb {F}=\mathbb {C}\) .
By Lemma REF 
there exist two mutually BJ orthogonal unit vectors \(x,y\in X\)  such that also \(\Phi (x)\ne 0\) . We identify the two-dimensional normed subspace \(\mathop {\mathrm {span}}_{\mathbb {C}}\lbrace x,y\rbrace \subseteq X\)  with the space \((\mathbb {C}^2,\Vert \cdot \Vert )\)  of row vectors in such a way that \(x,y\)  are identified with standard basis vectors, that is, \(x=(1,0)\)  and \(y=(0,1)\) . Recall that a \(\mathbb {C}\) -linear supporting functional at a point \((1,\alpha )\in \mathbb {C}^2\)  is given by
\(f_{\alpha }=\left(\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_1}}\,,\,\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_2}}\right)^\ast , \quad (z_k=x_k+i y_k).\) 
We have
\( \overline{ \left( \frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_k}} \right)} = \frac{\partial \overline{\Vert (1,\alpha )\Vert }}{\partial z_k} = \frac{\partial \Vert (1,\alpha )\Vert }{\partial z_k}\) 
for \(k = 1, 2\) , so the kernel of \(f_{\alpha }\)  contains a row vector
\(\left( \frac{\partial \Vert (1,\alpha )\Vert }{\partial z_2} \,,\, - {\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_1}}\right).\) 
Therefore, for every scalar \(\alpha \)  we get \((1,\alpha )\perp \left(\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_2} \,,\, - {\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_1}} \right)\) .
Since \(x,y\)  are mutually BJ orthogonal, their supporting functionals equal
\(\left(\frac{\partial \Vert (1,0)\Vert }{\partial \overline{z_1}},\frac{\partial \Vert (1,0)\Vert }{\partial \overline{z_2}}\right)^\ast =\mu _x\cdot (1,0)^\ast \; \hbox{ and } \left(\frac{\partial \Vert (0,1)\Vert }{\partial \overline{z_1}},\frac{\partial \Vert (0,1)\Vert }{\partial \overline{z_2}}\right)^\ast =\mu _y\cdot (0,1)^\ast ,\) 
for some nonzero \(\mu _x,\mu _y\in \mathbb {C}\) , respectively. Since partial derivatives of a smooth norm are continuous (see Rockafellar [1]}), the first supporting functional equals the limit of \(\left(\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_1}},\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_2}}\right)^\ast \)  as \(\alpha \in \mathbb {R}\)  goes to 0. Next, positive homogeneity of the norm implies \(\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_k}}=\frac{\partial \Vert (1/\alpha ,1)\Vert }{\partial \overline{z_k}}\)  (\(\alpha >0\) ), so the second supporting functional equals the limit of \(\left(\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_1}},\frac{\partial \Vert (1,\alpha )\Vert }{\partial \overline{z_2}}\right)^\ast \)  as \(\alpha \in \mathbb {R}\)  goes to \(\infty \) .
Hence, the vector
\(w_{\alpha }=\frac{\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_2}}{\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_1}}\alpha x - \alpha y\) 
is a well-defined function of \(\alpha \)  around \(\alpha =0\) , and is clearly parallel to (REF ), so \((x+\alpha y)\perp w_\alpha \) .
Also, its first component, i.e.,
\(\lambda (\alpha ):=\frac{\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_2}}{\frac{\partial \Vert (1,\alpha )\Vert }{\partial z_1}}\alpha ,\) 
is a continuous function which cannot be constantly zero because the numerator and the denominator are nonconstant functions and cannot vanish simultaneously.
As such, with \(\alpha \)  restricted to a suitable closed interval \(I=[0,\varepsilon ]\) , its range \(\Gamma \)  is a curve with more than one element in \(\mathbb {C}\) .
Thus, for every \(\alpha \in [0,\varepsilon ]\)  we have \((x+\alpha y)\perp (\lambda (\alpha ) x-\alpha y)\) . Since we assumed that \(\sigma \) -quasilinear \(\Phi \)  preserves BJ orthogonality, we get that
\(\Big (\Phi (x)+\sigma (\alpha )\Phi (y)\Big )\perp \Big (\sigma (\lambda (\alpha ))\Phi (x)-\sigma (\alpha )\Phi (y)\Big ).\) 
From here, and as \(x\perp y\)  imply \(\Phi (x)\perp \Phi (y)\) , the very definition of BJ orthogonality of \(\Phi (x)\)  and \(\Phi (y)\)  gives
\(\begin{aligned}\Vert \Phi (x)\Vert &\le \Vert \Phi (x)+\sigma (\alpha )\Phi (y)\Vert =\Vert \Phi (x+\alpha y)\Vert \\&\le \Vert \Phi (x+\alpha y)+\Phi (\lambda (\alpha ) x-\alpha y)\Vert =|1+\sigma (\lambda (\alpha ))|\cdot \Vert \Phi (x)\Vert \end{aligned}\) 
so that \(|\sigma (\mu )|\)  is bounded from below by 1 on the set \(\mu =1+\lambda (\alpha ) \in 1+\Gamma \subseteq \mathbb {C}\) . Being multiplicative, it follows that \(\sigma \)  is bounded from above on the reciprocal set \(\frac{1}{1+\Gamma }\) , which is also a curve with more than one element. The result then follows from Lemma REF .
